\(x^2+4x+5=2\sqrt{2x+3}\)

\(ĐK:x\ge-\dfrac{3}{2}\)

\(pt\Leftrightarrow(2x+3-2\sqrt{2x+3}+1)+x^2+2x+1=0\)

\(\Leftrightarrow\left(\sqrt{2x+3}-1\right)^2=-\left(x+1\right)^2\)

Vì \(\left(\sqrt{2x+3}-1\right)^2\ge0;-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}(\sqrt{2x+3}-1)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}=1\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)}\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy, pt có nghiệm duy nhất là x=-1

a: \(B=\dfrac{x^2+5x+5x+25}{x\left(x+5\right)}=\dfrac{x+5}{x}\)

A

B

A?

=>x-3=3

hay x=6

x=-18

x =18

x=-18

98

98

98

* x+y có thể bằng:
- 1+1
- 0+1
- 1+0

1+1
0+1
1+0

ta có: \(2x-1=2\left(x-3\right)+5\)

để \(2x-1⋮x-3\Rightarrow2\left(x-3\right)+5⋮x-3\\ m\text{à }x.nguy\text{ê}n\Rightarrow x-3nguy\text{ê}n\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{-5;5;1;-1\right\}\)

ta có bảng sau :

\(\Leftrightarrow2.\left(x-3\right)+5⋮x-3\)

\(do2.\left(x-3\right)⋮x-3\)

\(\Leftrightarrow5⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)

b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)